We consider the system fuhkop12rta1.

  Alphabet:

    app : [list * list] --> list 
    cons : [nat * list] --> list 
    hshuffle : [nat -> nat * list] --> list 
    nil : [] --> list 
    reverse : [list] --> list 

  Rules:

    app(nil, x) => x 
    app(cons(x, y), z) => cons(x, app(y, z)) 
    reverse(nil) => nil 
    reverse(cons(x, y)) => app(reverse(y), cons(x, nil)) 
    hshuffle(f, nil) => nil 
    hshuffle(f, cons(x, y)) => cons(f x, hshuffle(f, reverse(y))) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] app#(cons(X, Y), Z) =#> app#(Y, Z)   
    1] reverse#(cons(X, Y)) =#> app#(reverse(Y), cons(X, nil))   
    2] reverse#(cons(X, Y)) =#> reverse#(Y)   
    3] hshuffle#(F, cons(X, Y)) =#> hshuffle#(F, reverse(Y))   
    4] hshuffle#(F, cons(X, Y)) =#> reverse#(Y)   

  Rules R_0:

    app(nil, X) => X 
    app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
    reverse(nil) => nil 
    reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 
    hshuffle(F, nil) => nil 
    hshuffle(F, cons(X, Y)) => cons(F X, hshuffle(F, reverse(Y))) 

Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.

We consider the dependency pair problem (P_0, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  app#(cons(X, Y), Z) >? app#(Y, Z) 
  reverse#(cons(X, Y)) >? app#(reverse(Y), cons(X, nil)) 
  reverse#(cons(X, Y)) >? reverse#(Y) 
  hshuffle#(F, cons(X, Y)) >? hshuffle#(F, reverse(Y)) 
  hshuffle#(F, cons(X, Y)) >? reverse#(Y) 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  reverse(nil) >= nil 
  reverse(cons(X, Y)) >= app(reverse(Y), cons(X, nil)) 
  hshuffle(F, nil) >= nil 
  hshuffle(F, cons(X, Y)) >= cons(F X, hshuffle(F, reverse(Y))) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  app = \y0y1.2y1 
  app# = \y0y1.0 
  cons = \y0y1.0 
  hshuffle = \G0y1.G0(0) + 2G0(y1) 
  hshuffle# = \G0y1.3 
  nil = 0 
  reverse = \y0.0 
  reverse# = \y0.2 

Using this interpretation, the requirements translate to:

  [[app#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[app#(_x1, _x2)]] 
  [[reverse#(cons(_x0, _x1))]] = 2 > 0 = [[app#(reverse(_x1), cons(_x0, nil))]] 
  [[reverse#(cons(_x0, _x1))]] = 2 >= 2 = [[reverse#(_x1)]] 
  [[hshuffle#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[hshuffle#(_F0, reverse(_x2))]] 
  [[hshuffle#(_F0, cons(_x1, _x2))]] = 3 > 2 = [[reverse#(_x2)]] 
  [[app(nil, _x0)]] = 2x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 2x2 >= 0 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 0 >= 0 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[hshuffle(_F0, nil)]] = 3F0(0) >= 0 = [[nil]] 
  [[hshuffle(_F0, cons(_x1, _x2))]] = 3F0(0) >= 0 = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of:

  app#(cons(X, Y), Z) =#> app#(Y, Z)   
  reverse#(cons(X, Y)) =#> reverse#(Y)   
  hshuffle#(F, cons(X, Y)) =#> hshuffle#(F, reverse(Y))   

Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.

We consider the dependency pair problem (P_1, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  app#(cons(X, Y), Z) >? app#(Y, Z) 
  reverse#(cons(X, Y)) >? reverse#(Y) 
  hshuffle#(F, cons(X, Y)) >? hshuffle#(F, reverse(Y)) 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  reverse(nil) >= nil 
  reverse(cons(X, Y)) >= app(reverse(Y), cons(X, nil)) 
  hshuffle(F, nil) >= nil 
  hshuffle(F, cons(X, Y)) >= cons(F X, hshuffle(F, reverse(Y))) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  app = \y0y1.y0 + y1 
  app# = \y0y1.y0 
  cons = \y0y1.1 + y1 
  hshuffle = \G0y1.2 + 2y1 
  hshuffle# = \G0y1.0 
  nil = 0 
  reverse = \y0.y0 
  reverse# = \y0.0 

Using this interpretation, the requirements translate to:

  [[app#(cons(_x0, _x1), _x2)]] = 1 + x1 > x1 = [[app#(_x1, _x2)]] 
  [[reverse#(cons(_x0, _x1))]] = 0 >= 0 = [[reverse#(_x1)]] 
  [[hshuffle#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[hshuffle#(_F0, reverse(_x2))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[hshuffle(_F0, nil)]] = 2 >= 0 = [[nil]] 
  [[hshuffle(_F0, cons(_x1, _x2))]] = 4 + 2x2 >= 3 + 2x2 = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

  reverse#(cons(X, Y)) =#> reverse#(Y)   
  hshuffle#(F, cons(X, Y)) =#> hshuffle#(F, reverse(Y))   

Thus, the original system is terminating if (P_2, R_0, static, formative) is finite.

We consider the dependency pair problem (P_2, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  reverse#(cons(X, Y)) >? reverse#(Y) 
  hshuffle#(F, cons(X, Y)) >? hshuffle#(F, reverse(Y)) 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  reverse(nil) >= nil 
  reverse(cons(X, Y)) >= app(reverse(Y), cons(X, nil)) 
  hshuffle(F, nil) >= nil 
  hshuffle(F, cons(X, Y)) >= cons(F X, hshuffle(F, reverse(Y))) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  app = \y0y1.y0 + y1 
  cons = \y0y1.1 + y1 
  hshuffle = \G0y1.2y1 
  hshuffle# = \G0y1.0 
  nil = 0 
  reverse = \y0.y0 
  reverse# = \y0.y0 

Using this interpretation, the requirements translate to:

  [[reverse#(cons(_x0, _x1))]] = 1 + x1 > x1 = [[reverse#(_x1)]] 
  [[hshuffle#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[hshuffle#(_F0, reverse(_x2))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[hshuffle(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[hshuffle(_F0, cons(_x1, _x2))]] = 2 + 2x2 >= 1 + 2x2 = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of:

  hshuffle#(F, cons(X, Y)) =#> hshuffle#(F, reverse(Y))   

Thus, the original system is terminating if (P_3, R_0, static, formative) is finite.

We consider the dependency pair problem (P_3, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  hshuffle#(F, cons(X, Y)) >? hshuffle#(F, reverse(Y)) 
  app(nil, X) >= X 
  app(cons(X, Y), Z) >= cons(X, app(Y, Z)) 
  reverse(nil) >= nil 
  reverse(cons(X, Y)) >= app(reverse(Y), cons(X, nil)) 
  hshuffle(F, nil) >= nil 
  hshuffle(F, cons(X, Y)) >= cons(F X, hshuffle(F, reverse(Y))) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  app = \y0y1.y0 + y1 
  cons = \y0y1.1 + y1 
  hshuffle = \G0y1.2 + 2y1 
  hshuffle# = \G0y1.y1 
  nil = 0 
  reverse = \y0.y0 

Using this interpretation, the requirements translate to:

  [[hshuffle#(_F0, cons(_x1, _x2))]] = 1 + x2 > x2 = [[hshuffle#(_F0, reverse(_x2))]] 
  [[app(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[app(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 >= 1 + x1 + x2 = [[cons(_x0, app(_x1, _x2))]] 
  [[reverse(nil)]] = 0 >= 0 = [[nil]] 
  [[reverse(cons(_x0, _x1))]] = 1 + x1 >= 1 + x1 = [[app(reverse(_x1), cons(_x0, nil))]] 
  [[hshuffle(_F0, nil)]] = 2 >= 0 = [[nil]] 
  [[hshuffle(_F0, cons(_x1, _x2))]] = 4 + 2x2 >= 3 + 2x2 = [[cons(_F0 _x1, hshuffle(_F0, reverse(_x2)))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_3, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.